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$f : \mathbb{R} \to \mathbb{R}$ is continous function taking on values in $\mathbb{Z}$ or $\mathbb{Q}$. Show that it is constant function

Assume $f$ is not constant function.

Assume $f(c_1),f(c_2)$ such that both are rational (not equal to each other) . By density property of reals, a irrational number (say $r$) can be found in between them. By Intermediate value property , let $c$ be value on which $f(c)=r$. This is contradiction.

I am kind of stuck

Sophie Clad
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  • Where exactly are you stuck? This is the solution. You should be more confident in your solutions. – Mark Apr 19 '23 at 22:24

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I don't see why you're stuck. Looks fine to me. You might say "by the density of the irrational numbers" to be more specific than just "property."

A. Thomas Yerger
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